Question

A biologist wanted to estimate the mean length of western rattlesnakes. The following data (in inches) was collected: 40.2, 43.1, 45.5, 44.5, 39.5, 40.2, 41.0, 41.6, 43.1, 44.9 Find the 90% confidence interval for the mean length. ( Time saver: s = 2.158 ).

Answer 1

\mu = 41.1091, 43.6109

Explanation:

given data sets

40.2,43.1,45.5,44.5,39.5,40.2,41.0,41.6,43.1,44.9

sample mean = (sum of all value / number of value)

sample mean = 423.6/10 = 42.36

standard deviation = 2.158 given

t =1.833 for 90% confidence interval at df = n-1 = 9

90% confidence interval is given as

`\mu = mean length +\- t*\frac{s}{\sqrt {n}}`

`\mu = 42.36 +\- [1.833 * (2.158)/(\sqrt 10)]`

\mu = 41.1091, 43.6109